Expressing Complex Numbers in the Form x + iy | 将复数表示为 x + iy 的形式

📚 Expressing Complex Numbers in the Form x + iy | 将复数表示为 x + iy 的形式

In A-Level Mathematics, a complex number is usually introduced as an expression of the form \(x + iy\), where \(x\) and \(y\) are real numbers and \(i\) is the imaginary unit with \(i^2 = -1\). This article explains how to rewrite a wide range of complex expressions into this standard Cartesian form, which is essential for solving equations, performing arithmetic, and understanding geometric interpretations.

在 A-Level 数学中，复数通常被定义为形如 \(x + iy\) 的表达式，其中 \(x\) 和 \(y\) 是实数，\(i\) 是虚数单位，满足 \(i^2 = -1\)。本文讲解如何将各种复数表达式改写为这一标准笛卡尔形式，这是解方程、进行运算以及理解几何意义的关键。

1. Definition and Notation | 定义与记号

A complex number is written as \(z = x + iy\), where \(x = \operatorname{Re}(z)\) is the real part and \(y = \operatorname{Im}(z)\) is the imaginary part. Both \(x\) and \(y\) must be real numbers. The symbol \(i\) satisfies \(i^2 = -1\).

复数写作 \(z = x + iy\)，其中 \(x = \operatorname{Re}(z)\) 是实部，\(y = \operatorname{Im}(z)\) 是虚部。\(x\) 和 \(y\) 都必须是实数。符号 \(i\) 满足 \(i^2 = -1\)。

Example: \( 3 + 2i \) has real part 3 and imaginary part 2.
示例：\( 3 + 2i \) 的实部为 3，虚部为 2。
If \(y = 0\), the number is real; if \(x = 0\), it is purely imaginary.
若 \(y = 0\)，该数为实数；若 \(x = 0\)，则为纯虚数。

2. Adding and Subtracting Complex Numbers | 复数的加法与减法

To add or subtract complex numbers, combine their real parts and their imaginary parts separately. The result is already in the form \(x + iy\).

加减复数时，分别合并实部和虚部。结果自然就是 \(x + iy\) 的形式。

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) − (c + di) = (a − c) + (b − d)i

Example: \((2 + 3i) + (4 – 5i) = 6 – 2i\).
示例：\((2 + 3i) + (4 – 5i) = 6 – 2i\)。
Example: \((7 – i) – (2 + 3i) = 5 – 4i\).
示例：\((7 – i) – (2 + 3i) = 5 – 4i\)。

3. Multiplying Complex Numbers | 复数的乘法

Use the distributive law (FOIL) and replace \(i^2\) with \(-1\). Collect the real and imaginary terms to express the product in the form \(x + iy\).

使用分配律（FOIL）展开，并将 \(i^2\) 替换为 \(-1\)。合并实部和虚部项，即可将乘积表示为 \(x + iy\)。

(a + bi)(c + di) = (ac − bd) + (ad + bc)i

Example: \((2 + 3i)(1 – i) = 2 – 2i + 3i – 3i^2 = 2 + i + 3 = 5 + i\).
示例：\((2 + 3i)(1 – i) = 2 – 2i + 3i – 3i^2 = 2 + i + 3 = 5 + i\)。
Example: \((1 + i)^2 = 1 + 2i + i^2 = 2i\). Notice the real part is 0.
示例：\((1 + i)^2 = 1 + 2i + i^2 = 2i\)。注意实部为 0。

4. Complex Conjugate and Division | 共轭复数与除法

The conjugate of \(z = x + iy\) is \(\overline{z} = x – iy\). Multiplying a complex number by its conjugate gives a real number: \(z\overline{z} = x^2 + y^2\). To divide two complex numbers, multiply the numerator and denominator by the conjugate of the denominator.

复数 \(z = x + iy\) 的共轭为 \(\overline{z} = x – iy\)。复数与其共轭相乘得到实数：\(z\overline{z} = x^2 + y^2\)。两个复数相除时，将分子分母同时乘以分母的共轭。

\(\frac{a + bi}{c + di} = \frac{(a + bi)(c – di)}{(c + di)(c – di)} = \frac{ac + bd}{c^2 + d^2} + \frac{bc – ad}{c^2 + d^2}i\)

Example: \(\frac{1+i}{2-i} = \frac{(1+i)(2+i)}{(2-i)(2+i)} = \frac{2 + i + 2i + i^2}{5} = \frac{1 + 3i}{5} = \frac{1}{5} + \frac{3}{5}i\).
示例：\(\frac{1+i}{2-i} = \frac{(1+i)(2+i)}{(2-i)(2+i)} = \frac{2 + i + 2i + i^2}{5} = \frac{1 + 3i}{5} = \frac{1}{5} + \frac{3}{5}i\)。

5. Powers of i | i 的幂

Because \(i^2 = -1\), higher powers repeat in a cycle of length 4. To simplify \(i^n\), reduce \(n\) modulo 4.

因为 \(i^2 = -1\)，较高的幂以 4 为周期循环。要化简 \(i^n\)，将指数 \(n\) 对 4 取余。

\(n \bmod 4\)	0	1	2	3
\(i^n\)	1	\(i\)	−1	−\(i\)

Example: \(i^{27} = i^{24+3} = i^3 = -i\).
示例：\(i^{27} = i^{24+3} = i^3 = -i\)。
This pattern is useful when simplifying expressions involving powers of complex numbers.
这一规律在化简含复数幂的表达式时非常有用。

6. Solving Quadratic Equations | 解二次方程

When a quadratic equation has a negative discriminant, its solutions are complex conjugates. Using the quadratic formula and writing results in the form \(x + iy\) is a key skill.

当二次方程的判别式为负时，其解是共轭复数对。使用求根公式并将结果写成 \(x + iy\) 的形式是一项关键技能。

\(z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)

Example: Solve \(z^2 + 2z + 5 = 0\).
示例：解方程 \(z^2 + 2z + 5 = 0\)。
\(z = \frac{-2 \pm \sqrt{4 – 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i\).
\(z = \frac{-2 \pm \sqrt{4 – 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i\)。

7. Equating Real and Imaginary Parts | 实部与虚部相等

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. This principle allows us to convert a complex equation into two real equations.

两个复数相等当且仅当它们的实部相等且虚部相等。这一原理可以将一个复数方程转化为两个实数方程。

\(a + bi = c + di \iff a = c \text{ and } b = d\)

Example: If \((x + 2) + (y – 3)i = 4 + 5i\), then \(x + 2 = 4\) and \(y – 3 = 5\), so \(x = 2\), \(y = 8\).
示例：若 \((x + 2) + (y – 3)i = 4 + 5i\)，则 \(x + 2 = 4\)，\(y – 3 = 5\)，因此 \(x = 2\)，\(y = 8\)。
This method is frequently used to find unknown real constants in complex identities.
此方法常用于在复数恒等式中求未知实常数。

8. Expressing Rationalised Denominators | 有理化分母

An expression such as \(\frac{1}{a + bi}\) can be written in the form \(x + iy\) by multiplying numerator and denominator by the conjugate \(a – bi\).

形如 \(\frac{1}{a + bi}\) 的表达式可以通过将分子分母同时乘以共轭 \(a – bi\) 来写成 \(x + iy\)。

\(\frac{1}{a + bi} = \frac{a – bi}{a^2 + b^2} = \frac{a}{a^2 + b^2} – \frac{b}{a^2 + b^2}i\)

Example: \(\frac{1}{3 + 4i} = \frac{3 – 4i}{25} = \frac{3}{25} – \frac{4}{25}i\).
示例：\(\frac{1}{3 + 4i} = \frac{3 – 4i}{25} = \frac{3}{25} – \frac{4}{25}i\)。
Always simplify fully: both real and imaginary parts should be written as single fractions if required.
始终彻底化简：如果需要，实部和虚部都应写成单一分数形式。

9. Modulus and Argument in Cartesian Form | 模与辐角的笛卡尔形式

The modulus of \(z = x + iy\) is \(|z| = \sqrt{x^2 + y^2}\). The argument is \(\arg(z) = \tan^{-1}(y/x)\), adjusted for the correct quadrant. These are often used together to convert between polar and Cartesian forms.

复数 \(z = x + iy\) 的模为 \(|z| = \sqrt{x^2 + y^2}\)。辐角为 \(\arg(z) = \tan^{-1}(y/x)\)，需要根据象限调整。二者常用于极坐标与笛卡尔形式之间的转换。

Example: For \(z = 1 + i\), \(|z| = \sqrt{2}\) and \(\arg(z) = \pi/4\).
示例：对于 \(z = 1 + i\)，\(|z| = \sqrt{2}\)，\(\arg(z) = \pi/4\)。
If \(z = -1 + i\), \(|z| = \sqrt{2}\) but \(\arg(z) = 3\pi/4\), not \(-\pi/4\).
若 \(z = -1 + i\)，\(|z| = \sqrt{2}\)，但 \(\arg(z) = 3\pi/4\)，而不是 \(-\pi/4\)。

10. From Polar / Exponential Form to x + iy | 从极坐标或指数形式到 x + iy

Given a modulus \(r\) and argument \(\theta\), Euler’s formula \(e^{i\theta} = \cos\theta + i\sin\theta\) gives the Cartesian form directly:

给定模 \(r\) 和辐角 \(\theta\)，欧拉公式 \(e^{i\theta} = \cos\theta + i\sin\theta\) 可直接给出笛卡尔形式：

\(r e^{i\theta} = r\cos\theta + i r\sin\theta\)

Example: \(2e^{i\pi/3} = 2\cos(\pi/3) + i\,2\sin(\pi/3) = 1 + \sqrt{3}i\).
示例：\(2e^{i\pi/3} = 2\cos(\pi/3) + i\,2\sin(\pi/3) = 1 + \sqrt{3}i\)。
Remember to express the final answer with \(i\) written before the imaginary term, e.g. \(x + iy\), not \(x + yi\) unless clearly conventional.
注意最终答案通常写成 \(x + iy\)，虚部放在 \(i\) 后面，例如 \(x + iy\)，除非约定俗成也可写作 \(x + yi\)。

11. Square Roots of a Complex Number | 复数的平方根

To find \(\sqrt{a + bi}\) in the form \(x + iy\), set \((x + iy)^2 = a + bi\). Equating real and imaginary parts gives a system of equations for \(x\) and \(y\).

要求 \(\sqrt{a + bi}\) 并将其写成 \(x + iy\)，令 \((x + iy)^2 = a + bi\)。比较实部和虚部可得关于 \(x\) 和 \(y\) 的方程组。

\(x^2 – y^2 = a\), \(\quad 2xy = b\)

Also use \(x^2 + y^2 = \sqrt{a^2 + b^2}\) to help solve.
同时利用 \(x^2 + y^2 = \sqrt{a^2 + b^2}\) 来辅助求解。
Example: \((1 + i)^2 = 2i\), so \(\sqrt{2i} = 1 + i\) or \(-1 – i\).
示例：\((1 + i)^2 = 2i\)，所以 \(\sqrt{2i} = 1 + i\) 或 \(-1 – i\)。

12. Common Pitfalls and Exam Tips | 常见错误与考试提示

When writing complex numbers in the form \(x + iy\), students often make sign errors or forget to simplify \(i^2\). Practice rewriting expressions step by step.

在写成 \(x + iy\) 形式时，学生常犯符号错误或忘记化简 \(i^2\)。应逐步练习改写表达式。

Always check that your final answer has the structure (real number) + (real number)\(i\).
始终检查最终答案是否具有“实数 + 实数·\(i\)”的结构。
Never leave \(i\) in the denominator of a fraction in the final answer.
最终答案中切勿将 \(i\) 留在分母里。
Use the quadrant diagram when converting from polar to Cartesian form to determine the sign of \(\sin\theta\) and \(\cos\theta\).
从极坐标转换为笛卡尔形式时，使用象限图判断 \(\sin\theta\) 和 \(\cos\theta\) 的正负。
For AQA exam questions, always show the equating of real and imaginary parts when solving for unknowns.
对于 AQA 考试题目，在求解未知数时务必展示实部与虚部的比较过程。

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Expressing Complex Numbers in the Form x + iy | 将复数表示为 x + iy 的形式